Kp — equilibrium constant using partial pressures — Chemistry

Before this

This leads to

Kc and Kp at Pre-University · coming soon
Acid-base equilibria — Ka, Kb and pH · coming soon
Gibbs free energy, the criterion for spontaneity · coming soon

A pressure gauge cannot count molecules

In an ammonia plant, a gas mixture of nitrogen, hydrogen and ammonia flows through pipes at about 200 atmospheres. Nobody measures the concentration of each gas in mol/dm³. The instrument on the pipe is a pressure gauge, and it reads one number: the total pressure of the whole mixture.

The last lecture ended by saying that the same equilibrium constant appears for gases, built from partial pressures. This is that lecture. It answers two questions. How much of the total pressure belongs to each gas? And how do you write an equilibrium constant with those pressures? The answer is called K_p. By the end, you will also see exactly why squeezing a gas mixture changes its make-up, while K_p itself does not change at all.

Each gas pushes as if it were alone

A gas pushes on the walls of its container because its molecules keep hitting them. In an ideal gas, the molecules are tiny and almost never affect each other. So in a mixture, each gas hits the walls just as it would if the other gases were not there.

The partial pressure of a gas is the pressure it would exert if it filled the container on its own, at the same temperature. Because each gas pushes independently, the partial pressures add up to the total pressure:

P = p_{A} + p_{B} + p_{C} + \dots

John Dalton noticed this in 1801, while studying the water vapour in air, and published it in 1802. It is called Dalton's law of partial pressures. It is very close to exact at ordinary pressures, and only approximate at the very high pressures where molecules start to crowd each other.

How is the total shared out? The ideal gas equation, $p V = n RT$ , says that at a fixed volume and temperature, pressure is proportional to the number of moles. So each gas's share of the pressure equals its share of the molecules. That share is its mole fraction, written x (some books use the Greek letter χ, "chi"):

x_{A} = \frac{moles of A}{total moles of gas} p_{A} = x_{A} \times P

Pressure counts molecules. Each gas supplies the same share of the total pressure as its share of the molecules.

Mole fractions have no units, and they always add up to 1. That is why the partial pressures always add up to the total. In the figure, 2 molecules out of 10 are nitrogen, so nitrogen supplies 0.20 of the 200 atm: 40 atm.

Stop and think

A container holds 1 mol of hydrogen molecules and 1 mol of much heavier ammonia molecules. Does the ammonia supply more of the pressure, because its molecules are heavier?

Show the answer

No. Each gas supplies half of the pressure, because each is half of the molecules. Heavier molecules move more slowly at the same temperature, and the two effects cancel exactly: the average kinetic energy of a molecule depends only on the temperature. Pressure counts molecules, not mass.

Writing K_p

K_p has exactly the same shape as K_c. Products go on top and reactants underneath, each raised to the power of its coefficient. The only change is that each square-bracket concentration is replaced by an equilibrium partial pressure. For

a A(g) + b B(g) ⇌ c C(g) + d D(g)

The equilibrium constant K_p

$K_{p} = \frac{p _{C}^{c} p _{D}^{d}}{p _{A}^{a} p _{B}^{b}}$ , using equilibrium partial pressures.

Only gases go into K_p. At a fixed temperature, its value is the same for every equilibrium mixture.

Only gases have partial pressures, so solids and liquids are left out. For example, for $CaCO_{3} (s) ⇌ CaO(s) + CO_{2} (g)$ , the whole expression is $K_{p} = p_{CO_{2}}$ .

The units come from counting pressures, exactly as K_c's units came from counting concentrations. Use the pressure unit the question gives, usually atm or kPa (1 atm ≈ 101 kPa), and never mix two units in one calculation. If there are more pressures underneath than on top, the units are a negative power of the pressure unit. If the counts are equal, K_p has no units.

Now practise the most common route to a partial pressure: moles, then mole fractions, then partial pressures. One line in the working below is wrong. Find it.

Spot the error — partial pressures from moles

A student finds K_p for N₂ + 3H₂ ⇌ 2NH₃. The equilibrium mixture holds 1.0 mol N₂, 3.0 mol H₂ and 1.0 mol NH₃ at a total pressure of 100 atm. Tap the line that contains the error.

This working contains one error, in line 4.

Total moles = 1.0 + 3.0 + 1.0 = 5.0 mol. (Correct.)
x(N₂) = 1.0 ÷ 5.0 = 0.20, so p(N₂) = 0.20 × 100 = 20 atm. (Correct.)
x(NH₃) = 1.0 ÷ 5.0 = 0.20, so p(NH₃) = 20 atm. (Correct.)
p(H₂) = 3.0 × 100 = 300 atm. (Error: the moles were multiplied by P. The mole fraction is 3.0 ÷ 5.0 = 0.60, so p(H₂) = 60 atm. The partial pressures must add up to 100 atm.)
Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³). (Correct.)
Units: atm² ÷ atm⁴ = atm⁻². (Correct.)

With p(H₂) = 60 atm, K_p = 20² ÷ (20 × 60³) ≈ 9.3 × 10⁻⁵ atm⁻².

That error is the one examiners see most. A partial pressure needs the mole fraction, not the number of moles. Check every set of partial pressures by adding them: they must give the total pressure.

Why K_p is constant, and how it links to K_c

For each gas, the ideal gas equation gives $p_{A} = \frac{n _{A}}{V} RT$ . But $n_{A} / V$ is the concentration [A]. So every partial pressure is a concentration multiplied by the same number, RT:

p_{A} = [A] RT

Put this into K_p. Every factor of RT on top cancels one underneath. What is left over is RT raised to the power Δn, the change in the number of moles of gas:

K_{p} = K_{c} (RT)^{Δ n} Δ n = (c + d) - (a + b)

This is why K_p is a constant at all. K_c is fixed at a given temperature, and so is RT, so K_p is fixed too. It also shows when the two are equal. If the number of moles of gas does not change, Δn = 0, and K_p = K_c. For hydrogen and iodine forming hydrogen iodide, both are 50 at 443 °C.

One warning about R. With pressures in atm and concentrations in mol/dm³, use $R = 0.0821 dm^{3} atm K^{- 1} mol^{- 1}$ . The familiar value 8.31 J K⁻¹ mol⁻¹ works only with pressures in pascals and concentrations in mol/m³. T is always in kelvin.

Squeezing the mixture: Le Chatelier with numbers

This is the hard part of the lecture. Spend time here.

The Le Chatelier lecture predicted that raising the pressure on the Haber equilibrium shifts it towards ammonia, because 2 molecules of gas replace 4. K_p shows exactly how. Write each partial pressure as a mole fraction times P:

K_{p} = \frac{( x _{NH_{3}} P ) ^{2}}{( x _{N_{2}} P ) ( x _{H_{2}} P ) ^{3}} = \frac{x _{NH_{3}}^{2}}{x _{N_{2}} x _{H_{2}}^{3}} \times \frac{1}{P ^{2}}

K_p cannot change, because the temperature has not changed. So if P doubles, the factor $1/ P^{2}$ becomes a quarter as big. To keep K_p the same, the fraction made of mole fractions must become 4 times bigger. That can only happen if there is more ammonia and less nitrogen and hydrogen. The mixture shifts right.

That is it. That is the whole idea. Pressure does not change K_p. It changes the make-up of the mixture, so that K_p stays the same. Try it in the model below.

Explore — squeeze the Haber mixture (model)

The model keeps the temperature fixed, so K_p = 2.3 × 10⁻⁵ atm⁻², and nitrogen and hydrogen are always in the ratio 1 : 3. Change the total pressure and watch the equilibrium mixture and K_p.

Try:

Total pressure: 200 atm

This graph shows the ammonia mole fraction in the model's equilibrium mixture at each total pressure. K_p is the same, 2.3 × 10⁻⁵ atm⁻², at every point on the curve.

Ammonia mole fraction at equilibrium against total pressure, in the model (K_p = 2.3 × 10⁻⁵ atm⁻²).

P (atm)	x(NH₃)	x(N₂)	x(H₂)	p(NH₃) (atm)
100	0.121	0.220	0.659	12.1
200	0.200	0.200	0.600	40.0
400	0.303	0.174	0.522	121

The model assumes ideal gases and a mixture that reaches equilibrium. It leaves out the effect of very high pressure on real gases.

Notice two things. The ammonia fraction rises with pressure, but more and more slowly: doubling from 200 to 400 atm takes it only from 0.20 to about 0.30. And K_p stays the same at every pressure. Real ammonia plants run at about 200 atm and 400–450 °C, as a compromise between yield, rate and the cost of building equipment strong enough for higher pressures (Chemguide).

Worked examples

Worked Example 1 — K_p from mole fractions

For $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$ , an equilibrium mixture at 200 atm has mole fractions N₂ 0.20, H₂ 0.60 and NH₃ 0.20. Find K_p. (Numbers chosen for practice.)

Think first. Partial pressures first, then the expression, then the units.

$p_{N_{2}} = 0.20 \times 200 = 40$ atm, $p_{H_{2}} = 0.60 \times 200 = 120$ atm, $p_{NH_{3}} = 0.20 \times 200 = 40$ atm.

$K_{p} = \frac{4 0 ^{2}}{40 \times 12 0 ^{3}} = \frac{1600}{69 120 000} \approx 2.3 \times 1 0^{- 5} atm^{- 2}$ .

Units: two pressures on top and four underneath, so atm⁻². Δn = 2 − 4 = −2 agrees.

Check. 40 + 120 + 40 = 200 atm. ✓ This value describes the reaction at one temperature, at every pressure. It does not say "the mixture contains little ammonia at 200 atm": the mixture itself, with 20% ammonia, says that.

Worked Example 2 — K_p from moles

For $2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)$ , an equilibrium mixture contains 0.40 mol SO₂, 0.20 mol O₂ and 1.60 mol SO₃ at a total pressure of 5.0 atm. Find K_p. (Numbers chosen for practice.)

Think first. Total moles, mole fractions, partial pressures. Keep the unrounded values until the last line.

Total = 0.40 + 0.20 + 1.60 = 2.20 mol.

$p_{SO_{2}} = \frac{0.40}{2.20} \times 5.0 \approx 0.909$ atm, $p_{O_{2}} = \frac{0.20}{2.20} \times 5.0 \approx 0.455$ atm, $p_{SO_{3}} = \frac{1.60}{2.20} \times 5.0 \approx 3.636$ atm.

$K_{p} = \frac{p _{SO_{3}}^{2}}{p _{SO_{2}}^{2} p _{O_{2}}} \approx 35.2 atm^{- 1}$ (3 s.f., from the unrounded pressures).

Check. 0.909 + 0.455 + 3.636 = 5.000 atm. ✓ Working with the rounded pressures gives 35.1, which shows why rounding is left to the end.

Worked Example 3 — From a starting amount, then to K_c

1.00 mol of PCl₅ is heated to 250 °C in a container: $PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)$ . At equilibrium, 40% of it has dissociated and the total pressure is 2.00 atm. Find K_p, and then K_c. (Numbers chosen for practice.)

Think first. Use a start, change, equilibrium table, as in the last lecture.

Start: 1.00, 0, 0. Change: −0.40, +0.40, +0.40. Equilibrium: 0.60, 0.40, 0.40 mol, total 1.40 mol.

$p_{PCl_{5}} = \frac{0.60}{1.40} \times 2.00 \approx 0.857$ atm, and $p_{PCl_{3}} = p_{Cl_{2}} = \frac{0.40}{1.40} \times 2.00 \approx 0.571$ atm.

$K_{p} = \frac{0.571 \times 0.571}{0.857} \approx 0.381$ atm. Δn = 2 − 1 = +1, so the unit is atm.

T = 250 + 273 = 523 K, and $RT = 0.0821 \times 523 \approx 42.9$ . So $K_{c} = \frac{K _{p}}{RT} \approx \frac{0.381}{42.9} \approx 8.9 \times 1 0^{- 3}$ mol/dm³.

Check. Count phosphorus atoms: 0.60 + 0.40 = 1.00 mol, as at the start. ✓ Δn is positive, so K_p in atm is bigger than K_c in mol/dm³ here, because RT is bigger than 1 in these units.

Three traps

Trap 1 — Moles instead of mole fractions

A partial pressure is mole fraction × total pressure, not moles × total pressure. Add your partial pressures: if they do not give the total pressure, something is wrong.

Trap 2 — "Higher pressure increases K_p"

Only temperature changes K_p, as it does K_c. A change of pressure changes the mole fractions until the expression gives the same K_p again.

Trap 3 — The wrong R, or °C

In $K_{p} = K_{c} (RT)^{Δ n}$ , use 0.0821 dm³ atm K⁻¹ mol⁻¹ with atm and mol/dm³. The value 8.31 needs pascals and mol/m³. Temperature is always in kelvin.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A mixture contains 0.50 mol N₂, 1.50 mol H₂ and 0.25 mol NH₃ at a total pressure of 150 atm. It is at the same temperature as Worked Example 1, where K_p ≈ 2.3 × 10⁻⁵ atm⁻². Find each partial pressure and the value of the K_p expression for this mixture. Is the mixture at equilibrium? If not, which way does it react?

Hint 1
Total moles first: 0.50 + 1.50 + 0.25.

Hint 2
The expression for a mixture not yet at equilibrium is the reaction quotient Q, from the last lecture. Compare Q with K_p.

Full answer
Total = 2.25 mol. $p_{N_{2}} = \frac{0.50}{2.25} \times 150 \approx 33.3$ atm, $p_{H_{2}} = \frac{1.50}{2.25} \times 150 = 100$ atm, $p_{NH_{3}} = \frac{0.25}{2.25} \times 150 \approx 16.7$ atm. $Q = \frac{16. 7 ^{2}}{33.3 \times 10 0 ^{3}} \approx 8.3 \times 1 0^{- 6} atm^{- 2}$ . Q is smaller than K_p, so the mixture is not at equilibrium. It reacts to the right, making more ammonia. Check: 33.3 + 100 + 16.7 = 150 atm. ✓
Write K_p, with units, for (a) $CaCO_{3} (s) ⇌ CaO(s) + CO_{2} (g)$ ; (b) $N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)$ ; (c) $H_{2} (g) + I_{2} (g) ⇌ 2 HI(g)$ . Pressures in kPa.

Hint 1
Leave out solids. Then count pressures on top and underneath.

Hint 2
(a) has one pressure on top and none underneath.

Full answer
(a) $K_{p} = p_{CO_{2}}$ , in kPa. (b) $K_{p} = \frac{p _{NO_{2}}^{2}}{p _{N_{2} O_{4}}}$ , in kPa. (c) $K_{p} = \frac{p _{HI}^{2}}{p _{H_{2}} p _{I_{2}}}$ , no units. Check with Δn: (a) +1, (b) +1, (c) 0, which match the units. ✓
For hydrogen and iodine at 443 °C, K_c = 50. Find K_p. Then explain why squeezing this mixture into half the volume does not change its make-up.

Hint 1
Find Δn first.

Hint 2
Write K_p with mole fractions and P, as for the Haber reaction. What power of P is left?

Full answer
Δn = 2 − 2 = 0, so $K_{p} = K_{c} (RT)^{0} = 50$ , with no units. Written with mole fractions, $K_{p} = \frac{x _{HI}^{2} P ^{2}}{x _{H_{2}} P \times x _{I_{2}} P} = \frac{x _{HI}^{2}}{x _{H_{2}} x _{I_{2}}}$ . P cancels completely, so a change of pressure gives the mole fractions no reason to change. Check with Le Chatelier: 2 molecules of gas on each side, so pressure has nothing to oppose. ✓
A stretch. In the Haber model of this lecture (K_p ≈ 2.3 × 10⁻⁵ atm⁻², nitrogen and hydrogen always in the ratio 1 : 3), the ammonia mole fraction is 0.20 at 200 atm. (a) Show that the mole-fraction part of the expression, $x_{NH_{3}}^{2} \div (x_{N_{2}} x_{H_{2}}^{3})$ , is about 0.93 at 200 atm. (b) At 400 atm, what must it become? (c) Check that an ammonia fraction of 0.30 gives roughly that value.

Hint 1
At 200 atm the mole fractions are 0.20, 0.20 and 0.60. For (b), the factor 1/P² becomes a quarter.

Hint 2
For (c), if $x_{NH_{3}} = 0.30$ , the remaining 0.70 is shared 1 : 3, so $x_{N_{2}} = 0.175$ and $x_{H_{2}} = 0.525$ .

Full answer
(a) $0.2 0^{2} \div (0.20 \times 0.6 0^{3}) = 0.04 \div 0.0432 \approx 0.93$ . (b) Doubling P makes 1/P² a quarter, so the mole-fraction part must become 4 times bigger: about 3.7. (c) $0.3 0^{2} \div (0.175 \times 0.52 5^{3}) = 0.09 \div 0.0253 \approx 3.6$ , close to 3.7. The exact value from the model is 0.303. Check: K_p = 3.6 ÷ 400² ≈ 2.2 × 10⁻⁵ atm⁻², about the same as at 200 atm. ✓ The ammonia fraction rose by half, not doubled: Le Chatelier's "opposes, not undoes", in numbers.

The bigger idea

K_c and K_p are two windows onto the same equilibrium. Concentration is the natural measure in a solution, and pressure is the natural measure for a gas, because a pressure gauge is what you can read. The ideal gas equation links the two, and that is why one formula, $K_{p} = K_{c} (RT)^{Δ n}$ , carries you from one to the other.

There is a deeper reason why gas equilibria are written with pressures. At Pre-University level, equilibrium constants are linked to energy through the Gibbs free energy, and that link is written with pressures compared with a standard pressure of 100 kPa. Every "standard" value in the energetics lectures, written with ⦵, is measured against that same pressure. K_p is the first step towards that connection between how far a reaction goes and how much energy it can release.

Key concepts in this lesson

The partial pressure of a gas is the pressure it would exert alone in the same container, and the partial pressures add up to the total pressure (Dalton's law).
Each partial pressure is the gas's mole fraction times the total pressure, $p_{A} = x_{A} P$ , because pressure counts molecules.
K_p has the same shape as K_c, with equilibrium partial pressures of gases only, and units found by counting pressures.
$K_{p} = K_{c} (RT)^{Δ n}$ , so the two are equal when the number of moles of gas does not change.
Only temperature changes K_p; a change of pressure changes the mole fractions, which is Le Chatelier's principle shown in numbers.